Bai2 Bang Hoang Huy 2011 38€¦ · acta mathematica vietnamica 145 volume 36, number 2, 2011, pp. 145–167 some properties of orlicz-lorentz spaces ha huy bang, nguyen van hoang

ACTA MATHEMATICA VIETNAMICA 145Volume 36, Number 2, 2011, pp. 145–167

SOME PROPERTIES OF ORLICZ-LORENTZ SPACES

HA HUY BANG, NGUYEN VAN HOANG AND VU NHAT HUY

Dedicated to Tran Duc Van on the occasion of his sixtieth birthday

Abstract. In this paper we study some fundamental properties of Orlicz-Lorentz spaces defined on R such as finding their dual spaces, giving bestconstants for the inequalities between the Orlicz norm and the Luxemburgnorm on Orlicz-Lorentz spaces and establishing the Kolmogorov inequality inthese spaces.

1. Orlicz-Lorentz spaces

Orlicz-Lorentz spaces as a generalization of Orlicz spaces Lϕ and Lorentz spacesΛω have been studied by many authors (we refer to [9, 10, 11, 12, 14, 18, 19] forbasic properties of Orlicz- Lorentz spaces as well to the references therein). In thispaper we study some fundamental properties of Orlicz-Lorentz spaces defined onthe real line ΛR

ϕ,ω. We first find the dual spaces of ΛRϕ,ω. Note that the dual spaces

of Orlicz-Lorentz spaces defined on (0,+∞) or (0, 1) were studied in [11]. Nextwe introduce the Orlicz norm on ΛR

ϕ,ω which defined by using the MRϕ∗,ω space

and then we give a simple formula to calculate the Orlicz norm directly by ϕ,ω.On Orlicz spaces, it is known that the Orlicz norm and the Luxemburg norm areequivalent, and it will be shown that the corresponding norms on Orlicz-Lorentzspaces are also equivalent. Moreover, we investigate best constants C1, C2 for theinequalities between the Orlicz norm and the Luxemburg norm on Orlicz-Lorentzspaces and we notice that these results for the special case when ΛR

ϕ,ω becomesOrlicz spaces will be published in [2]. The dual equality between the Orlicz normon Orlicz-Lorentz spaces and the norm on MR

ϕ∗,ω is also given. Finally, we provethe Kolmogorov inequality in the Orlicz-Lorentz spaces.

Let us first recall some notations of Orlicz- Lorentz spaces:Let (Ω, µ) := (Ω,Σ, µ) be a measure space with the complete and σ-finite measureµ, L0(µ) be a space of all µ-equivalent classes of Σ-measurable functions on Ωwith topology of the convergence in measure on µ-finite sets.A Banach space (E, ‖.‖E) is called the Banach function space on (Ω, µ) if it isa subspace of L0(µ), and there exists a function h ∈ E such that h > 0 a.e. on

Received May 9, 2011.2000 Mathematics Subject Classification. 46E30.Key words and phrases. Orlicz- Lorentz spaces.This work was supported by the Vietnam National Foundation for Science and Technology

Development (Project N0 101.01.50.09).

146 HA HUY BANG, NGUYEN VAN HOANG AND VU NHAT HUY

Ω and if f ∈ L0(µ), g ∈ E and |f | ≤ |g| a.e. on Ω then f ∈ E and we have‖f‖E ≤ ‖g‖E . Moreover, if the unit ball BE = f ∈ E : ‖f‖E ≤ 1 is closed onL0(µ), then we say that E has the Fatou property. A Banach function space Eis said to be symmetric if for every f ∈ L0(µ) and g ∈ E such that µf = µg, thenf ∈ E and ‖f‖E = ‖g‖E , where for any h ∈ L0(µ), µh denotes the distributionof h, defined by

µh(t) = µ(x ∈ Ω : |h(x)| > t), t ≥ 0.

Let E be a Banach function space on (Ω, µ). Then the Kothe dual space E′

of E is a Banach function space, which can be identified with the space of allfunctionals possessing an integral representation, that is,

E′= g ∈ L0(µ) : ‖g‖E′ = sup

‖f‖E≤1

∫

Ω

|fg|dµ <∞.

Given ϕ : [0,∞) → [0,∞) an Orlicz function (i.e., it is a convex function, takesvalue zero only at zero) and ω : (0,∞) → (0,∞) a weight function (i.e., it is

a non-increasing function and locally integrable and∞∫

0

ωdx = ∞). The Orlicz -

Lorentz space ΛΩϕ,ω on (Ω, µ) is the set of all functions f(x) ∈ L0(µ) such that

∞∫

0

ϕ(λf∗(x))ω(x)dx <∞

for some λ > 0, where f∗ is the non-increasing rearrangement of f defined by

f∗(x) = infλ > 0 : µf (λ) ≤ x,

with x > 0 (by convention, inf ∅ = ∞).It is easy to check that ΛΩ

ϕ,ω is a symmetric Banach function space, with theFatou property, equipped with the Luxemburg norm

‖f‖ΛΩϕ,ω

= infλ > 0 :

∞∫

0

ϕ(f∗(x)

λ)ω(x)dx ≤ 1.

Note that: If ω ≡ 1 then ΛΩϕ,ω is the Orlicz function space LΩ

ϕ ; if ϕ(t) = t then

ΛΩϕ,ω is the Lorentz function space ΛΩ

ω .Recall that ϕ is an N-function if lim

t→0ϕ(t)/t = 0 and lim

t→+∞ϕ(t)/t = +∞; the

Orlicz function ϕ satisfies ∆2-condition (we write, ϕ ∈ ∆2) if there exists C > 0such that ϕ(2t) ≤ Cϕ(t) ∀t > 0; the Orlicz function ϕ : [0,+∞) 7−→ [0,+∞)satisfies the ∇2-condition (we write, ϕ ∈ ∇2) if there exists a number l > 1 suchthat ϕ(x) ≤ 1

2lϕ(lx) ∀x ≥ 0. We easily have the following remarks:

Remark 1.1.

(i) If f(x) ∈ E′and 0 ≤ fn ↑ |f | then lim

n→+∞‖fn‖E

′ = ‖f‖E′ ;

(ii) If f(x), fn(x), n = 1, 2, ... are measurable functions satisfying |fn| ↑ |f | thenf∗n ↑ f∗;

SOME PROPERTIES OF ORLICZ-LORENTZ SPACES 147

(iii) If f(x), g(x) are measurable functions then

∫

Ω

|f(x)g(x)|dµ ≤

+∞∫

0

f∗(x)g∗(x)dx.

Remark 1.2. Let ϕ be an N-function. Then the three following conditions areequivalent:(i) ϕ ∈ ∇2;(ii) There exists β > 1 such that xψ(x) > βϕ(x) ∀x > 0, where ψ(x) is the leftderivative of ϕ;(iii) There are the numbers l > 1 and δl > 0 such that ϕ(lx) ≥ (l+δl)ϕ(x)∀x > 0.

Denote by ϕ∗ the Young conjugate function of ϕ, that is

ϕ∗(t) = supst− ϕ(s)|s ≥ 0, t ≥ 0.

Then we have the following resultYoung’s inequality. Let ϕ be an N-function. Then

xy ≤ ϕ(x) + ϕ∗(y) ∀x, y ≥ 0

and it becomes the equality if and only if y ∈ [ψ(x), η(x)], where ψ, η are the leftand the right derivatives of ϕ.We define

I(f) =

∞∫

0

ϕ∗(f∗(x)

ω(x))ω(x)dx

for any f(x) ∈ L0(µ) and

MΩϕ∗,ω = f(x) ∈ L0(µ) : I(

f

λ) <∞ with some λ > 0.

In the space MΩϕ∗,ω we define a monotone and homogeneous functional

‖f‖MΩϕ∗,ω

= infλ > 0 : I(f

λ) ≤ 1.

Put

S(t) =

t∫

0

ω(s)ds, t > 0,

we call the weight function ω regular if there is a constant K > 1 such thatS(2t) ≥ KS(t) for any t > 0. It is easy to prove that ω is regular if and only ifthere exists C > 0 such that tω(t) ≤ S(t) ≤ Ctω(t) for any t > 0.Let f(x), g(x) be two positive functions, we write f g if there exist C1, C2 > 0such that C1f(x) ≤ g(x) ≤ C2f(x). Put I = (0,+∞).It was proved in [11] thatTheorem A. Let ω be a weight function and ϕ(t) = t or ϕ be an N-function.Then the following assertions are true:

(i) If ω is regular, then (ΛIϕ,ω)

′= M I

ϕ∗,ω and ‖.‖(ΛIϕ,ω)′‖.‖MI

ϕ∗,ω;


(ii) If ϕ ∈ ∆2 and (ΛIϕ,ω)

′= M I

ϕ∗,ω, then ω is regular.

Theorem A shows the relation between Orlicz-Lorentz spaces and M Iϕ∗,ω, that

is, the Kothe dual space of Orlicz-Lorentz space (ΛIϕ,ω) is the M I

ϕ∗,ω with someconditions of ϕ,ω.

2. Main results

We state the following theorem as an extension of Theorem A

Theorem 2.1. Let ω be a weight function and ϕ(t) = t or ϕ be an N-function.Then the following assertions are true

(i) If ω is regular, then (ΛRϕ,ω)

′= MR

ϕ∗,ω and ‖.‖(ΛRϕ,ω)′‖.‖MR

ϕ∗,ω;

(ii) If ϕ ∈ ∆2 and (ΛRϕ,ω)

′= MR

ϕ∗,ω, then ω is regular.

To prove Theorem 2.1, we need the following lemmas.

Lemma 2.2. Let Ω = R or I and f(x) be a measurable function. Then thefollowing are true

(i) f ∈ ΛΩϕ,ω if and only if f∗ ∈ ΛI

ϕ,ω, and we have ‖f‖ΛΩϕ,ω

= ‖f∗‖ΛIϕ,ω

;

(ii) f ∈MΩϕ∗,ω if and only if f∗ ∈M I

ϕ∗,ω, and we have ‖f‖MΩϕ∗,ω

= ‖f∗‖MIϕ∗,ω

;

(iii) f ∈ (ΛΩϕ,ω)

′if and only if f∗ ∈ (ΛI

ϕ,ω)′, and we have ‖f‖(ΛΩ

ϕ,ω)′ = ‖f∗‖(ΛIϕ,ω)′ .

Proof. (i) and (ii) is evident from their definitions. Let us prove (iii). Using (i)

and Remark 1.1, we have if f∗(x) ∈ (ΛIϕ,ω)

′, then f(x) ∈ (ΛΩ

ϕ,ω)′and

‖f‖(ΛΩϕ,ω)′ ≤ ‖f∗‖(ΛI

ϕ,ω)′ .

Conversely, suppose that f(x) ∈ (ΛΩϕ,ω)

′, we have to prove f∗(x) ∈ (ΛI

ϕ,ω)′

and‖f∗‖(ΛI

ϕ,ω)′ ≤ ‖f‖(ΛΩϕ,ω)′ . By Remark 1.1, we only prove for f(x) being a non-

negative, simple function on Ω. So f∗(x) is a nonnegative simple function on I,too. For any simple function g(x) ∈ ΛI

ϕ,ω satisfying ‖g‖ΛIϕ,ω

≤ 1, there is a simple

function h(x) on Ω such that h∗(x) = g∗(x) and∫

Ω

|f(x)h(x)|dµ =

∫

I

f∗(x)g∗(x)dx.

Hence, h(x) ∈ ΛΩϕ,ω and ‖h‖ΛΩ

ϕ,ω≤ 1, and then

∫

I

|f∗(x)g(x)|dx ≤

∫

I

f∗(x)g∗(x)dx =

∫

Ω

|f(x)h(x)|dx ≤ ‖f‖(ΛΩϕ,ω)′ .

If g(x) is an arbitrary function in ΛIϕ,ω such that ‖g‖ΛI

ϕ,ω≤ 1, there is a sequence

gn(x) of nonnegative simple functions on I such that gn(x) ↑ |g(x)|. So gn(x) ∈ΛI

ϕ,ω and ‖gn‖ΛIϕ,ω

≤ 1. By the monotone convergence theorem, we have∫

I

|f∗(x)g(x)|dx = limn→∞

∫

I

|f∗(x)gn(x)|dx ≤ ‖f‖(ΛΩϕ,ω)

′ .


Hence, f∗(x) ∈ (ΛIϕ,ω)

′and ‖f∗‖(ΛI

ϕ,ω)′ ≤ ‖f‖(ΛΩϕ,ω)′ . The proof is complete.

Lemma 2.3. Let ϕ be an Orlicz function, ω be a weight function and supposethat (ΛR

ϕ,ω)′= MR

ϕ∗,ω. Then there is a constant K > 0 satisfying

‖g‖MRϕ∗ ,ω

≤ K‖g‖(ΛRϕ,ω)′ ∀g(x) ∈ (ΛR

ϕ,ω)′.

Proof. This lemma is proved similarly as in the proof of Lemma 2 in [11].

Proof of Theorem 2.1. (i). By the regularity of ω, we have (ΛIϕ,ω)

′= M I

ϕ∗,ω and

2−1‖f‖(ΛIϕ,ω)′ ≤ ‖f‖MI

ϕ∗,ω≤ 4C‖f‖(ΛI

ϕ,ω)′

where C is a constant (see Theorem 2 (i) [11]). It follows from Lemma 2.2

that f(x) ∈ (ΛRϕ,ω)

′⇐⇒ f∗ ∈ (ΛI

ϕ,ω)′

= M Iϕ∗,ω ⇐⇒ f(x) ∈ MR

ϕ∗,ω. Therefore,

(ΛRϕ,ω)

′= MR

ϕ∗,ω, and

‖f‖(ΛRϕ,ω)′ = ‖f∗‖(ΛI

ϕ,ω)′ ≤ 2‖f∗‖MIϕ∗,ω

= 2‖f‖MRϕ∗,ω

,

‖f‖MRϕ∗,ω

= ‖f∗‖MIϕ∗,ω

≤ 4C‖f∗‖(ΛIϕ,ω)′ = 4C‖f‖(ΛR

ϕ,ω)′ .

i.e., ‖.‖(ΛRϕ,ω)′‖.‖MR

ϕ∗,ω.

By Theorem 2(ii) [11], we will prove that (ΛIϕ,ω)

′= M I

ϕ∗,ω. From Young’s in-

equality we have M Iϕ∗,ω ⊂ (ΛI

ϕ,ω)′, so we only have to prove (ΛI

ϕ,ω)′⊂M I

ϕ∗,ω.

Given f ∈ (ΛIϕ,ω)

′, there is a sequence fn of nonnegative simple functions on I

such that fn ↑ |f |. Hence, fn ∈ (ΛIϕ,ω)

′for all n ∈ N and ‖fn‖(ΛI

ϕ,ω)′ ↑ ‖f‖(ΛI

ϕ,ω)′ .

We choose a sequence hn of simple functions on R such that h∗n = f∗n for

all n ∈ N. Thus hn ∈ (ΛRϕ,ω)

′and ‖hn‖(ΛR

ϕ,ω)′ = ‖fn‖(ΛIϕ,ω)′ . By assumption

(ΛRϕ,ω)

′= MR

ϕ∗,ω, we obtain hn ∈ MRϕ∗,ω. Lemma 2.2(ii) yields f∗n = h∗n ∈ M I

ϕ∗,ω

and ‖f∗n‖MIϕ∗,ω

= ‖h∗n‖MIϕ∗,ω

= ‖hn‖MRϕ∗,ω

. From Lemma 2.3, there is a constant

K > 0 such that

‖g‖MRϕ∗ ,ω

≤ K‖g‖(ΛRϕ,ω)′ ∀g ∈ (ΛR

ϕ,ω)′.

Hence

‖f∗n‖MIϕ∗,ω

= ‖hn‖MRϕ∗,ω

≤ K‖hn‖(ΛRϕ,ω)′ = K‖fn‖(ΛI

ϕ,ω)′ ≤ K‖f‖(ΛIϕ,ω)′ .

So supn≥1

‖f∗n‖MIϕ∗,ω

<∞. Then it follows from f∗n ↑ f∗ that f∗ ∈M Iϕ∗,ω. Therefore,

f ∈M Iϕ∗,ω. That is (ΛI

ϕ,ω)′⊂M I

ϕ∗,ω. The proof is complete.

Next, we study the norms on Orlicz-Lorentz spaces. We know that ΛΩϕ,ω is the

Banach space with the Luxemburg norm defined by

‖f‖ΛΩϕ,ω

= infλ > 0 :

∫

I

ϕ(f∗(x)

λ)ω(x)dx ≤ 1.


On this space we define another norm called the Orlicz norm as follows

‖f‖1ΛΩ

ϕ,ω= sup

∫

Ω

|f(x)g(x)|dµ : ‖g‖MΩϕ∗ ,ω

≤ 1.(1)

Note that if ω ≡ 1 then ΛΩϕ,ω is the usual Orlicz function space LΩ

ϕ and the

Luxemburg norm, the Orlicz norm on ΛΩϕ,ω become respectively the Luxemburg

norm, the Orlicz norm on LΩϕ which can be seen in [19].

From the definition of the Orlicz norm, we have∫

Ω

|f(x)g(x)|dµ ≤ ‖f‖1ΛΩ

ϕ,ω‖g‖MΩ

ϕ∗ ,ω∀f(x) ∈ ΛΩ

ϕ,ω, g(x) ∈MΩϕ∗,ω.

In the following theorem, we give a formula to compute the Orlicz norm.

Theorem 2.4. Let ϕ be an N-function and f(x) ∈ ΛRϕ,ω. Then

‖f‖1ΛR

ϕ,ω= inf

1

k(1 +

∫

I

ϕ(kf∗(x))ω(x)dx) : k > 0.(2)

To obtain Theorem 2.4, the following result is important.

Lemma 2.5. Let ϕ be an Orlicz function, ψ be the left derivative of ϕ and thefunction f(x) ∈ ΛR

ϕ,ω satisfy

∫

I

ϕ∗(ψ(k0f∗(x))ω(x)dx = 1(3)

for some k0. Then we have

‖f‖1ΛR

ϕ,ω=

∫

I

f∗(x)ψ(k0f∗(x))ω(x)dx =

1

k0(1 +

∫

I

ϕ(k0f∗(x))ω(x)dx).

Proof. From (1) and (3), we have

‖f‖1ΛR

ϕ,ω= sup

∫

R

|f(x)g(x)|dx : ‖g‖MRϕ∗ ,ω

≤ 1(4)

≤1

k0(1 +

∫

I

ϕ(k0f∗(x))ω(x)dx)

=1

k0(

∫

I

ϕ∗(ψ(k0f∗(x))ω(x)dx +

∫

I

ϕ(k0f∗(x))ω(x)dx)

=

∫

I

f∗(x)ψ(k0f∗(x))ω(x)dx.


We define g(x) = ω(x)ψ(k0f∗(x)), x > 0. Then ‖g‖MI

ϕ∗ ,ω≤ 1. Therefore, it

follows from the definition of the Orlicz norm that

‖f‖1ΛR

ϕ,ω= ‖f∗‖1

ΛIϕ,ω

≥

∫

I

f∗(x)g(x)dx =

∫

I

f∗(x)ψ(k0f∗(x))ω(x)dx.(5)

From (4) and (5), the proof is complete.

Proof of Theorem 2.4. For k > 0 and g(x) ∈ MRϕ∗,ω satisfying ‖g‖MR

ϕ∗,ω≤ 1, we

have∫

I

ϕ∗(g∗(x)

ω(x))ω(x)dx ≤ 1.

Thus∫

R

|f(x)g(x)|dx ≤

∫

I

f∗(x)g∗(x)dx =1

k

∫

I

(g∗(x)

ω(x))(kf∗(x))ω(x)dx

≤1

k(

∫

I

ϕ∗(g∗(x)

ω(x))ω(x)dx+

∫

I

ϕ(kf∗(x))ω(x)dx)

≤1

k(1 +

∫

I

ϕ(kf∗(x))ω(x)dx).

This inequality implies that

‖f‖1ΛR

ϕ,ω≤ inf

1

k(1 +

∫

I

ϕ(kf∗(x))ω(x)dx) : k > 0.(6)

Next, we will prove the opposite inequality: Let us first suppose that ψ is acontinuous function. There exists a sequence fn(x) of nonnegative, simplefunctions such that fn(x) ↑ |f(x)| a.e. Hence, ‖fn‖

1Λϕ,ω

↑ ‖f‖1Λϕ,ω

, and f∗n(x) ↑

f∗(x). We have m(x ∈ R : fn(x) 6= 0) <∞ for all n ∈ N and ϕ∗ is a continuousfunction. So there is a sequence kn of positive numbers such that

∫

I

ϕ∗(ψ(knf∗n(x)))ω(x)dx = 1 ∀n ∈ N.

Therefore, it follows from Lemma 2.5 that

kn‖fn‖1ΛR

ϕ,ω=

∫

I

ϕ(knf∗(x))ω(x)dx + 1 ∀n ∈ N.(7)

Because f∗n(x) is an increasing sequence of functions and kn is a decreasingsequence, there exists the limit lim

n→∞kn = k∗, and then from (7), we have

k∗‖f‖1ΛR

ϕ,ω= lim inf

n→∞

∫

I

ϕ(knf∗(x))ω(x)dx + 1 ≥

∫

I

ϕ(k∗f∗(x))ω(x)dx + 1.


Therefore, k∗ > 0 and

‖f‖1ΛR

ϕ,ω≥

1

k∗(

∫

I

ϕ(k∗f∗(x))ω(x)dx + 1)

≥ inf1

k(1 +

∫

I

ϕ(kf∗(x))ω(x)dx) : k > 0.

From this and (6), we get

‖f‖1ΛR

ϕ,ω= inf

1

k(1 +

∫

I

ϕ(kf∗(x))ω(x)dx) : k > 0.

If ψ is not a continuous function, it is known that ψ is left continuous. For allε ∈ (0, 1), we approximate ψ by a nondecreasing, continuous ψ1 such that

ϕ((1 − ε)x) ≤ ϕ1(x) ≤ ϕ(x) ∀x > 0,

where ϕ1 is a convex function having ψ1 as the left derivative. It is easy to provethat ϕ1 is an N-function, ΛR

ϕ,ω = ΛRϕ1,ω and

(1 − ε)‖f‖1ΛR

ϕ,ω≤ ‖f‖1

ΛRϕ1,ω

≤ ‖f‖1ΛR

ϕ,ω.

Hence, due to the result proved for continuous functions , we have

‖f‖1ΛR

ϕ,ω≥ ‖f‖1

ΛRϕ1,ω

= inf1

k(1 +

∫

I

ϕ1(kf∗(x))ω(x)dx : k > 0

≥ inf1

k(1 +

∫

I

ϕ((1 − ε)kf∗(x))ω(x)dx : k > 0

= (1 − ε) inf1

k(1 +

∫

I

ϕ(kf∗(x))ω(x)dx : k > 0.

Letting ε→ 0, we get

‖f‖1ΛR

ϕ,ω≥ inf

1

k(1 +

∫

I

ϕ(kf∗(x))ω(x)dx : k > 0.

Combining this equation with (6) we obtain

‖f‖1ΛR

ϕ,ω= inf

1

k(1 +

∫

I

ϕ(kf∗(x))ω(x)dx : k > 0.

The proof is complete.

We will show, in particular, that on Orlicz-Lorentz spaces the Orlicz norm andthe Luxemburg norm are equivalent , and satisfy the following inequalities

‖f‖ΛRϕ,ω

≤ ‖f‖1ΛR

ϕ,ω≤ 2‖f‖ΛR

ϕ,ω∀f ∈ ΛR

ϕ,ω.


Moreover, we find the best constants for the inequalities between these norms.Suppose that C1 is the largest number and C2 is the smallest number such that

C1‖f‖ΛRϕ,ω

≤ ‖f‖1

Λ|Rϕ,ω

≤ C2‖f‖ΛRϕ,ω

∀f ∈ ΛR

ϕ,ω.(8)

It is well known that the Orlicz norm has the Fatou property, that is, if 0 ≤ fn ≤f ∈ ΛR

ϕ,ω, then ‖fn‖ΛRϕ,ω

→ ‖f‖ΛRϕ,ω

whenever fn → f a.e. Hence, condition (8)

is equivalent to the following

C1‖f‖ΛRϕ,ω

≤ ‖f‖1ΛR

ϕ,ω≤ C2‖f‖ΛR

ϕ,ω(9)

for all functions f(x) ∈ ΛRϕ,ω which are nonnegative, simple and satisfying

‖f‖ΛRϕ,ω

= 1. From the above definition, we have 1 ≤ C1 ≤ C2 ≤ 2. Put

H(k) = supt>0

ϕ(kt)

ϕ(t), D(k) = inf

t>0

ϕ(kt)

ϕ(t).

Clearly, the functions D(k),H(k) are increasing, D(k) ≤ H(k) ≤ k for any0 ≤ k ≤ 1 and k ≤ D(k) ≤ H(k) for any k > 1. From now on, we denote by f−1

the inverse function of f .

Theorem 2.6. Let ϕ be an Orlicz function. Then(i) If ϕ is an N-function, we have

C1 = infc>0

1

cϕ−1∗ (c)ϕ−1(c) = inf

k>0

1 +D(k)

k;(10)

(ii) We always have

C2 ≤ infk>0

1 +H(k)

k,(11)

and if ϕ is an N-function then

supc>0

1

cϕ−1∗ (c)ϕ−1(c) ≤ C2.(12)

Proof. (i) We have

infk>0

1

k(1 +D(k)) = inf

k>0

1

k

(

1 + infx>0

ϕ(kx)

ϕ(x)

)

= infk>0

1

k(1 + inf

c>0

ϕ(kϕ−1(c))

c)(13)

= infc>0

1

cinfk>0

ϕ∗(ϕ−1∗ (c) + ϕ(kϕ−1(c))

kc= inf

c>0

1

cϕ−1(c)ϕ−1

∗ (c).

If f(x) ∈ ΛRϕ,ω is a simple function such that ‖f‖ΛR

ϕ,ω= 1, then

∫

I

ϕ(f∗(x))ω(x)dx = 1.


Therefore, it follows from Theorem 2.4 that

‖f‖1ΛR

ϕ,ω= inf

1

k

(

1 +

∫

I

ϕ(kf∗(x))ω(x) dx)

: k > 0

≥ inf1

k

(

1 +D(k)

∫

I

ϕ(f∗(x))ω(x) dx)

: k > 0

≥ infk>0

1 +D(k)

k.

Therefore, due to (9), we have

C1 ≥ infk>0

1 +D(k)

k.(14)

For any c > 0, we choose t > 0 such thatt∫

0

ω(x)dx = 1/c. We put f(x) = χ(0,t)(x).

By an immediate computation, we have

‖f‖ΛRϕ,ω

=1

ϕ−1(c)and ‖f‖1

ΛRϕ,ω

= infk>0

1

k(1 + ϕ(k)c) =

1

cϕ−1∗ (c).

Hence

C1 ≤‖f‖1

ΛRϕ,ω

‖f‖ΛRϕ,ω

=1

cϕ−1∗ (c)ϕ−1(c) ∀c > 0.

This implies

C1 ≤ infc>0

1

cϕ−1(c)ϕ−1

∗ (c).(15)

Combining (13), (14) and (15), we obtain (10).(ii) Let f ∈ ΛR

ϕ,ω and ‖f‖1ΛR

ϕ,ω≤ 1. Then it follows from Theorem 2.4 that

‖f‖1ΛR

ϕ,ω≤

1

k(1 +

∫

I

ϕ(kf∗(x))ω(x)dx)

≤1

k(1 +H(k)

∫

I

ϕ(f∗(x))ω(x)dx) ≤1 +H(k)

k∀k > 0,

which gives C2 ≤ infk>0

1+H(k)k .

Similarly as above for C1, we get

C2 ≥ supc>0

1

cϕ−1(c)ϕ−1

∗ (c),

if ϕ is an N-function. The proof is complete.

Now we find the conditions so that C1 = 1.

Theorem 2.7. Let ϕ be an N-function. Then C1 > 1 if and only if ϕ ∈ ∆2∩∇2.


Proof. Necessity. Assume C1 > 1. We have to prove that ϕ ∈ ∆2 ∩∇2. Indeed,assume the contrary that ϕ /∈ ∆2 ∩∇2. Then ϕ /∈ ∆2 or ϕ /∈ ∇2. From Theorem2.6, we have

C1 = inft>0

1 +D(t)

t.

If ϕ /∈ ∆2, there exists a sequence of positive numbers xn such that ϕ(xn) ≥nϕ(xn/2) ∀n ∈ N. Fix t ∈ (0, 1) and choose n0 ∈ N such that 1/2 ≥ tn0. Thenfor all n > n0 we have ϕ(xn) ≥ nϕ(xn/2) ≥ nϕ(tn0xn). So, it follows fromϕ(tn0xn) ≥ (D(t))n0ϕ(xn) that 1 ≥ n(D(t))n0 ∀n > n0, and then D(t) = 0 forall t ∈ (0, 1). Hence

C1 ≤ inft∈(0,1)

1 +D(t)

t= inf

t∈(0,1)

1

t= 1.

Therefore, by C1 ≥ 1 we have C1 = 1.If ϕ /∈ ∇2, it follows from Remark 1.2 that for any t > 1, for all δ > 0 there existsx > 0 such that

ϕ(tx) < (t+ δ)ϕ(x).

Therefore,

D(t) = infx>0

ϕ(tx)

ϕ(x)≤ t+ δ.

Letting δ → 0, we obtain D(t) = t ∀t > 1. So we have

C1 ≤ inft>1

1 +D(t)

t= inf

t>1

1 + t

t= 1.

From this inequality and by C1 ≥ 1, we get C1 = 1, which contradicts C1 > 1.So, ϕ ∈ ∆2 ∩∇2 has been proved.Sufficiency. Assume ϕ ∈ ∆2 ∩ ∇2, we have to show C1 > 1. Indeed, sinceϕ ∈ ∆2, D(1/2) > 0. Since ϕ ∈ ∇2, there exists β > 1 such that

xψ(x)

ϕ(x)> β ∀x > 0,

where ψ is the left derivative of ϕ (see (ii) in Remark 1.2). Therefore, for all t > 1we have

lnϕ(tx)

ϕ(x)=

tx∫

x

ψ(y)

ϕ(y)dy ≥

tx∫

x

β

ydy = β ln t ∀x > 0.

This implies D(t) ≥ tβ. Hence

inft≥1

1 +D(t)

t≥ inf

t>1

1 + tβ

t> 1.

Then it follows from

inf1>t≥1/2

1 +D(t)

t≥ inf

1>t≥1/2(1 +D(t)) ≥ 1 +D(1/2) > 1


and

inf1/2≥t>0

1 +D(t)

t≥ 2,

that

C1 = inft>0

1 +D(t)

t> 1.

The proof is complete.

We have the following result:

Lemma 2.8. Let ϕ be an Orlicz function with continuous left derivative ψ. Put

H(k) = supx>0

ϕ(kx)

ϕ(x), a = sup

x>0

xψ(x)

ϕ(x), b = inf

x>0

xψ(x)

ϕ(x).

Then H has the left derivative and the right derivative at 1 and H′

+(1) = a,

H′

−(1) = b.

Proof. For k > 1, x > 0 we have

lnϕ(kx)

ϕ(x)=

kx∫

x

ψ(t)

ϕ(t)dt ≤

kx∫

x

a

tdt = a ln k.

Thus H(k) ≤ ka. Hence

lim supk→1+

H(k) −H(1)

k − 1≤ lim

k→1+

ka − 1

k − 1= a.(16)

Otherwise, let c ∈ (0, a). There exist x0 > 0, δ > 0 such that

xψ(x)

ϕ(x)> c ∀x ∈ (x0, x0 + δ).

For k ∈ (1, 1 + δx0

), we have (x0, kx0) ⊂ (x0, x0 + δ), and then

lnϕ(kx0)

ϕ(x0)=

kx0∫

x0

ψ(t)

ϕ(t)dt ≥

kx0∫

x0

c

tdt = c ln k.

This implies that

H(k) ≥ϕ(kx0)

ϕ(x0)≥ kc.

Hence

lim infk→1+

H(k) − 1

k − 1≥ lim

k→1+

kc − 1

k − 1= c.

Letting c → a and using (16), we see that H has the right derivative at 1 and

H′

+(1) = a. Next, we prove that H′

−(1) = b. Indeed, for k < 1 we have

lnϕ(x)

ϕ(kx)=

x∫

kx

ψ(t)

ϕ(t)dt ≥

x∫

kx

b

tdt = −b ln k = − ln kb ∀x > 0,


which gives H(k) ≤ kb. Hence

lim infk→1−

1 −H(k)

1 − k≥ lim

k→1−

1 − kb

1 − k= b.(17)

On the other hand, for d > b, there exists x0 > 0 satisfying

x0ψ(x0)

ϕ(x0)< d,

and then there exists δ > 0 such that

xψ(x)

ϕ(x)< d ∀x ∈ (x0 − δ, x0).

For 1 − δx0< k < 1 we get (kx0, x0) ⊂ (x0 − δ, x0), and then

lnϕ(x0)

ϕ(kx0)=

x0∫

kx0

ψ(t)

ϕ(t)dt ≤

x0∫

kx0

d

tdt = − ln kd.

It follows that

H(k) ≥ϕ(kx0)

ϕ(x0)≥ kd ∀k ∈ (1 −

δ

x0, 1).

Therefore,

lim supk→1−

1 −H(k)

1 − k≤ lim

k→1+

1 − kd

1 − k= d ∀d > b.(18)

Combining (17) and (18), we obtain that H has the left derivative at 1 and

H′

−(1) = b. The proof is complete.

Theorem 2.9. Let ϕ be an Orlicz function and its left derivative ψ be continuous.Then C2 = 2 if and only if

infx>0

xψ(x)

ϕ(x)≤ 2 ≤ sup

x>0

xψ(x)

ϕ(x).(19)

Proof. Necessary. Assume that C2 = 2, we have to show (19) . Indeed, put g(k) =(1 + H(k))/k. Then g(1) = 2 and due to Theorem 2.6, we get C2 ≤ infg(k) :k > 0. So, g(1)= ming(k) : k > 0. Since H has the left derivative and theright derivative at 1, g also has these derivatives at 1. Moreover, it follows fromg(t) ≥ g(1) ∀t > 0 that g

′

+(1) ≥ 0 ≥ g′

−(1). Thus

H′

+(1) ≥ 2 ≥ H′

−(1).

From this and using Lemma 2.8, we have (19).Sufficiency. Assume that (19) is true, we need to prove C2 = 2. Indeed, for allε ∈ (0, 1), by the continuity of ψ, there exists x0 > 0 such that

x0ψ(x0)

ϕ(x0)∈ (2 − ε, 2 + ε).

Putf(x) = x0χ(0,t)(x), g(x) = ψ(x0)ω(x)χ(0,t)(x),


where t is chosen such that ϕ(x0)t∫

0

ω(x)dx = 1 − ε. Hence

∫

R

ϕ(|f(x)|)ω(x)dx = 1 − ε

and

∫

R

f(x)g(x)dx =

t∫

0

x0ψ(x0)ω(x)dx

=x0ψ(x0)

ϕ(x0)

t∫

0

ϕ(x0)ω(x)dx ∈ ((2 − ε)(1 − ε), (2 + ε)(1 − ε)).

Thus

2 − 3ε <

∫

R

f(x)g(x)dx < 2 − ε.

Using Young’s equality, we get∫

R

f(x)g(x)dx =

∫

I

ϕ(f∗(x))ω(x)dx +

∫

I

ϕ∗(g∗(x)

ω(x))ω(x)dx.

Then it follows from∫

I ϕ(|f∗(x)|)ω(x)dx = 1 − ε that∫

I

ϕ∗(g∗(x)

ω(x))ω(x)dx ≤ 1.

So, we obtain

‖f‖ΛRϕ,ω

≤ 1, ‖g‖MRϕ∗ ,ω

≤ 1 and

∫

R

f(x)g(x)dx > 2 − 3ε.

Hence

C2 ≥‖f‖1

ΛRϕ,ω

‖f‖ΛRϕ,ω

≥

∫

R

f(x)g(x)dx > 2 − 3ε.

Letting ε→ 0 we get C2 ≥ 2. So, C2 = 2. The proof is complete.

Theorem 2.10. Let ϕ be an Orlicz function. For each g(x) ∈MRϕ∗,ω, we define

‖g‖1MR

ϕ∗ ,ω= sup

∫

R

|f(x)g(x)|dx : ‖f‖ΛRϕ,ω

≤ 1.(20)

Then we have the following dual equality

‖f‖ΛRϕ,ω

= sup

∫

R

|f(x)g(x)|dx : ‖g‖1MR

ϕ∗ ,ω≤ 1.(21)


Proof. From (20), we obtain the following inequality

∫

R

|f(x)g(x)|dx ≤ ‖f‖ΛRϕ,ω

‖g‖1MR

ϕ∗,ω.

Therefore,

‖f‖ΛRϕ,ω

≥ sup

∫

R


ϕ∗ ,ω≤ 1.(22)

Next, we prove the inverse inequality

‖f‖ΛRϕ,ω

≤ sup

∫

R


ϕ∗ ,ω≤ 1.(23)

We can assume that ‖f‖ΛRϕ,ω

= 1.

If f(x) is a simple function, then for any ε > 0 we have

∫

I

ϕ((1 + ε)f∗(x))ω(x)dx > 1.

Put

g(x) =ψ((1 + ε)f∗(x))ω(x)

1 +∫

I

ϕ∗(ψ((1 + ε)f∗(x))ω(x)dxχ(0,∞)

(g(x) is well-defined because ψ(f∗(x)) is a simple function too, so we have∫

I

ϕ∗(ψ((1 + ε)f∗(x))ω(x)dx <∞). Using Young’s inequality, we have

‖g‖1MR

ϕ∗,ω= sup

∫

R

|h(x)g(x)|dx : ‖h‖ΛRϕ,ω

≤ 1

≤

∫

I

ψ((1 + ε)f∗(x))h∗(x)ω(x)dx

1 +∫

I

ϕ∗(ψ((1 + ε)f∗(x))ω(x)dx

≤

∫

I

ψ((1 + ε)f∗(x))h∗(x)ω(x)dx

∫

I

ϕ(h∗(x))ω(x)dx +∫

I

ϕ∗(ψ((1 + ε)f∗(x)))ω(x)dx≤ 1.


Thus we get

sup

∫

R

|f(x)h(x)|dx : ‖h‖1MR

ϕ∗,ω≤ 1

≥

∫

I

f∗(x)g∗(x)dx =

∫

I

ψ((1 + ε)f∗(x))f∗(x)ω(x)dx

1 +∫

I

ϕ∗(ψ((1 + ε)f∗(x))ω(x)dx

=1

1 + ε

∫

I

ϕ((1 + ε)f∗(x))ω(x)dx +∫

I

ϕ∗(ψ((1 + ε)f∗(x)))ω(x)dx

1 +∫

I

ϕ∗(ψ((1 + ε)f∗(x))ω(x)dx

≥1

1 + ε∀ε > 0.

Hence

sup

∫

R

|f(x)h(x)|dx : ‖h‖1MR

ϕ∗,ω≤ 1 ≥ 1 = ‖f‖ΛR

ϕ,ω.

Therefore, (23) is true for simple functions f(x) .If f(x) is an arbitrary function, then approximating f(x) by a sequence of simplefunctions we get (23). Combining (22) with (23), we have (21). The proof iscomplete.

3. The Kolmogorov inequality in Orlicz-Lorentz space

The Landau-Kolmogorov inequality

‖f (k)‖n∞ ≤ K(k, n)‖f‖n−k

∞ ‖f (n)‖k∞,(24)

where 0 < k < n, is well known and has many interesting applications and gener-alizations (see [1, 3, 4, 5, 6, 7, 20, 21, 22, 23]). Its study was initiated by Landau[17] and Hadamard [8] (the case n = 2). For functions on the whole real line R,Kolmogorov [15] succeeded in finding in explicit form the best possible constantsK(k, n) = Ck,n in (24), and Stein proved in [22] that inequality (24) still holdsfor Lp-norm, 1 ≤ p < ∞, with these constants (the same situation also happensfor an arbitrary Orlicz norm [1]). In this section will prove that the Kolmogorovinequality still holds for the Orlicz norm and the Luxemburg norm in Orlicz-Lorentz spaces. For simplicity of notations, we we denote ΛR

ϕ,ω by Λϕ,ω, ‖.‖ΛRϕ,ω

by ‖.‖Λϕ,ω , ‖.‖1ΛR

ϕ,ωby ‖.‖ϕ,ω, MR

ϕ∗,ω by Mϕ∗,ω and ‖.‖MRϕ∗,ω

by ‖.‖Mϕ∗,ω . Note

that if ω is regular (that is∫ t0 ω(s)ds tω(t)), then Mϕ∗,ω is a linear space, and

‖ · ‖Mϕ∗,ω is a quasi-norm. Especially, Λϕ,ω ⊂ S′(R) is the space of all tempered

generalized functions this follow from the fact that∫ t0 ω(s)ds tω(t).

We have the following lemmas:

Lemma 3.1. Let f ∈ Λϕ,ω and g ∈ L1(R). Then f ∗ g ∈ Λϕ,ω and

‖f ∗ g‖ϕ,ω ≤ ‖f‖ϕ,ω‖g‖1


and

‖f ∗ g‖Λϕ,ω ≤ ‖f‖Λϕ,ω‖g‖1.

Lemma 3.2. Let n ≥ 1. If f ∈ L1,loc(R) and its generalized nth derivative

g ∈ L1,loc(R), then f can be redefined on a set of measure zero so that f (n−1) is

absolutely continuous and f (n) = g a.e. on R.

Now, we state our theorem.

Theorem 3.3. Let ϕ be an arbitrary Orlicz function, ω be a weight function, fand its generalized derivative f (n) be in Λϕ,ω. Then f (k) ∈ Λϕ,ω for all 0 < k < nand

‖f (k)‖nϕ,ω ≤ Ck,n‖f‖

n−kϕ,ω ‖f (n)‖k

ϕ,ω ,(25)

where Ck,n are the best constants defined in the Kolmogorov inequality (for thecase p = ∞).

Proof. We begin to prove (25) with the assumption that f (k) ∈ Λϕ,ω, with k =0, 1, ..., n. Indeed, fix 0 < k < n and let ε > 0 be given. We choose a functionvε ∈Mϕ∗,ω, ‖vε‖Mϕ∗,ω ≤ 1 such that

∣

∣

∫

R

f (k)(x)vε(x)dx∣

∣ ≥ ‖f (k)‖ϕ,ω − ε.(26)

Put

Fε(x) =

∫

R

f(x+ y)vε(y)dy.

Then Fε(x) ∈ L∞(R) by the definition of the Orlicz norm, and

F (r)ε (x) =

∫

R

f (r)(x+ y)vε(y)dy, 0 ≤ r ≤ n(27)


in the distribution sense. Actually, for every function ψ(x) ∈ C∞0 (R) it follows

from the assumption and the definition of the Orlicz norm that

〈F (r)ε (x), ψ(x)〉 = (−1)r〈Fε(x), ψ

(r)(x)〉

= (−1)r∫

R

∫

R

f(x+ y)vε(y)dy

ψ(r)(x)dx

= (−1)r∫

R

vε(y)

∫

R

f(x+ y)ψ(r)(x)dx

dy

=

∫

R

vε(y)

∫

R

f (r)(x+ y)ψ(x)dx

dy

=

∫

R

∫

R

f (r)(x+ y)vε(y)dy

ψ(x)dx

= 〈

∫

R

f (r)(x+ y)vε(y)dy , ψ(x)〉.

So we have proved (27).Since ‖vε‖Mϕ∗,ω ≤ 1, clearly, for all x ∈ R,

|F (r)ε (x)| ≤ ‖f (r)(x+ ·)‖ϕ,ω‖vε‖Mϕ∗,ω ≤ ‖f (r)‖ϕ,ω.

Now, we prove the continuity of F(r)ε on R. Indeed, put h(x) = f (r)(x), ht(x) =

f (r)(x + t), g(x) = vε(x). So, to prove the continuity of F(r)ε on R we only have

to show that

limt→0

∫

R

(ht(x) − h(x))g(x)dx = 0.(28)

To do this, it is sufficient to prove for real nonnegative value functions h(x). Since

h ∈ Λϕ,ω and g ∈ (Λϕ,ω)′, we have

∞∫

0

h∗(x)g∗(x)dx <∞.(29)

We first prove (28) whenever h(x) = χA(x) is the characteristic function of themeasurable set A, there are two cases, that isCase 1: m(A) < +∞. We denote A− t := x − t : x ∈ A, Ct := A∆(A− t).Then it follows from m(A) < ∞ that lim

t→0m(Ct) = lim

t→0m(A∆(A − t)) = 0. We


have h∗(x) = χ(0,m(A))(x). From this and (29) we get

m(A)∫

0

g∗(x)dx < +∞.

Hence, for any ε > 0, there exists δ > 0 such thatδ∫

0

g∗(x)dx < ε, and then there

is t0 > 0 such that m(A∆(A− t)) < δ for all |t| < t0. Therefore, for |t| < t0 :∣

∣

∣

∫

R

(χA(x+ t) − χA(x))g(x)dx∣

∣

∣ ≤

∫

R

∣

∣

∣(χA(x+ t) − χA(x))g(x)∣

∣

∣dx

=

∫

Ct

∣

∣

∣χCt(x)∣

∣

∣.∣

∣

∣g(x)∣

∣

∣dx ≤

m(Ct)∫

0

g∗(x)dx < ε.

That is

limt→0

∫

R

(χA(x+ t) − χA(x))g(x)dx = 0.

Case 2: m(A) = +∞. Then h∗(x) ≡ 1 on I. Therefore, from (29), we see thatg∗(x) is integrable on I. Thus, g(x) ∈ L1(R), and then lim

t→0‖g − g−t‖L1(R) = 0.

Therefore, it follows from∫

R

(χA(x+ t) − χA(x))g(x)dx =

∫

R

χA(x)g(x − t)dx−

∫

R

χA(x)g(x)dx

=

∫

R

χA(x)(g(x − t) − g(x))dx

≤

∫

R

|(g(x − t) − g(x))|dx = ‖g−t − g‖L1(R)

that

limt→0

∫

R

(χA(x+ t) − χA(x))g(x)dx = 0,

i.e., (28) is true for h(x) = χA(x) being the characteristic function of the mea-surable set A.By the linearity of integral, (28) is true for all simple functions h(x) satisfyingthe condition of the theorem.If h(x) is a nonnegative, measurable function, we consider the sequence of func-tions hn(x)∞n=1 as follows

hn(x) =

n2n−1∑

k=0

k

2nχAn,k

(x) + nχAn(x),


where An,k = x : k2n ≤ h(x) < k+1

2n and An = x : h(x) ≥ n. Then it iseasy to check that hn(x) ↑ h(x) a.e., and lim

n→∞m(An) = 0. Given ε > 0 and

δ > 0. We choose n0 such that 1/2n < ε and m(An) < δ for all n ≥ n0, thenx : h(x) − hn(x) ≥ ε ⊂ An. Hence

m(x : |h(x) − hn(x)| ≥ ε) ≤ m(An) < δ.

That is hnm→ f . So (hn − h)∗(x) → 0. By Lebesgue’s dominated convergence

theorem, we obtain

limn→∞

∫

I

(hn − h)∗(x)g∗(x)dx = 0.

Then it follows from∣

∣

∣

∫

R

(h(x+ t)−h(x))g(x)dx∣

∣

∣ =∣

∣

∣

∫

R

(h(x+ t) − hn(x+ t))g(x)dx

+

∫

R

(hn(x+ t) − hn(x))g(x)dx +

∫

R

(hn(x) − h(x))g(x)dx∣

∣

∣

≤ 2

∫

I

(hn − h)∗(x)g∗(x)dx+∣

∣

∣

∫

R

(hn(x+ t) − hn(x))g(x)dx∣

∣

∣

that

lim supt→0

∣

∣

∣

∫

R

(h(x + t) − h(x))g(x)dx∣

∣

∣≤ 2

∞∫

0

(hn − h)∗(x)g∗(x)dx ∀n ∈ N.

Hence

lim supt→0

∣

∣

∣

∫

R

(h(x+ t) − h(x))g(x)dx∣

∣

∣ ≤ 2 limn→∞

∞∫

0

(hn − h)∗(x)g∗(x)dx = 0.

This gives

limt→0

∫

R

(h(x+ t) − h(x))g(x)dx = 0.

So, (28) has been proved.

The functions F(r)ε are continuous and bounded on R, therefore it follows from

the Landau-Kolmogorov inequality and (26)-(27) that(

‖f (k)‖ϕ,ω − ε)n

≤ |F (k)ε (0)|n ≤ ‖F (k)

ε ‖n∞ ≤ Ck,n‖Fε‖

n−k∞ ‖F (n)

ε ‖k∞.

On the other hand,

‖Fε‖∞ ≤ ‖f(x+ ·)‖ϕ,ω‖vε(·)‖Mϕ∗ ,ω ≤ ‖f‖ϕ,ω ,

‖F (n)ε ‖∞ ≤ ‖f (n)(x+ ·)‖ϕ,ω‖vε(·)‖Mϕ∗ ,ω ≤ ‖f (n)‖ϕ,ω .

Hence(

‖f (k)‖ϕ,ω − ε)n

≤ Ck,n‖f‖n−kϕ,ω ‖f (n)‖k

ϕ,ω.


By letting ε→ 0, we have (25).

To complete the proof, it remains to show that f (k) ∈ Λϕ,ω, with 1 ≤ k ≤ n−1

if f, f (n) ∈ Λϕ,ω. Indeed, by Lemma 3.2 we can assume that f, f ′, . . . , f (n−1) are

continuous on R and f (n−1) is absolutely continuous on R, because f, f (n) ∈ Λϕ,ω.Let ψ ∈ C∞

0 (R), ψ ≥ 0, ψ(x) = 0 for |x| ≥ 1 and∫

R

ψ(x)dx = 1. We put

ψλ(x) = 1/λψ(x/λ), λ > 0 and fλ = f ∗ ψλ. Then fλ ∈ C∞(R) and f(k)λ =

f ∗ψ(k)λ = f (k) ∗ψλ, k ≥ 0. It follows from Lemma 3.1 that f

(k)λ ∈ Λϕ,ω. Then by

the fact proved above, we obtain

‖f(k)λ ‖n

ϕ,ω ≤ Ck,n‖fλ‖n−kϕ,ω ‖f

(n)λ ‖k

ϕ,ω, 0 < k < n.

It follows from the following inequalities

‖fλ‖ϕ,ω ≤ ‖f‖ϕ,ω‖ψλ‖1 = ‖f‖ϕ,ω,

‖f(n)λ ‖ϕ,ω ≤ ‖f (n)‖ϕ,ω‖ψλ‖1 = ‖f (n)‖ϕ,ω

that the set f(k)λ λ∈R+

is bounded in Λϕ,ω and, by the continuity of f(k)λ , lim

λ→0f

(k)λ (x)

= limλ→0

f (k) ∗ ψλ = f (k)(x)∀x ∈ R. Indeed, for all x ∈ R, we have

|f(k)λ (x) − f (k)(x)| =

∣

∣

∣

∣

∫

R

(

f (k)(x− y) − f (k)(y))

ψλ(y)dy

∣

∣

∣

∣

≤

∫

|λ|≤ε

∣

∣f (k)(x− y) − f (k)(y)∣

∣ψλ(y)dy

≤ sup|y|≤λ

∣

∣f (k)(x− y) − f (k)(y)∣

∣ → 0 as λ→ 0+.

Put gλ(x) = inf0<µ≤λ

|f(k)µ (x)|. Then gλ ∈ Λϕ,ω for all λ ∈ R+, the set gλλ∈R+

is bounded in Λϕ,ω and gλ ↑ |f (k)| as λ → 0+. Therefore, g∗λ ↑ f (k)∗ as λ → 0+.Choose M > 0 such that ‖gλ‖ < M for all λ ∈ R+. So,

∞∫

0

ϕ(g∗λ(t)

M)ω(t)dt ≤ 1 ∀λ ∈ R+.

Letting λ→ 0+, by the monotone convergence theorem, we get

∞∫

0

ϕ(f (k)∗(t)

M)ω(t)dt ≤ 1.

Hence f (k) ∈ Λϕ,ω for 1 ≤ k ≤ n− 1. The proof is complete.

From the proof of (28) we have the following result.


Proposition 3.4. Let f, g be measurable functions satisfying the following con-dition

∞∫

0

f∗(x)g∗(x)dx < +∞.

Then

limt→0

∫

R

(f(x+ t) − f(x))g(x)dx = 0.

From Proposition 3.4 , we have the following:

Corollary 3.5. Let f ∈ ΛRϕ,ω, g ∈MR

ϕ∗,ω. Then

limt→0

∫

R

(f(x+ t) − f(x))g(x)dx = 0.

For the Luxemburg norm ‖ · ‖Λϕ,ω , the Kolmogorov inequality also holds:

Theorem 3.6. Let ϕ be an arbitrary Orlicz function, ω be a weight function, fand its generalized derivative f (n) be in Λϕ,ω. Then f (k) ∈ Λϕ,ω for all 0 < k < nand

‖f (k)‖nΛϕ,ω

≤ Ck,n‖f‖n−kΛϕ,ω

‖f (n)‖kΛϕ,ω

.

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Institute of Mathematics,

18 Hoang Quoc Viet Street, Cau Giay, Hanoi, Vietnam

E-mail address: [email protected]

Institute of Mathematics,

18 Hoang Quoc Viet Street, Cau Giay, Hanoi, Vietnam

E-mail address: [email protected]

Department of Mathematics,

College of Science, Vietnam National University,

334 Nguyen Trai Street, Thanh Xuan, Hanoi, Vietnam

E-mail address: nhat [email protected]

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Bai2 Bang Hoang Huy 2011 38€¦ · acta mathematica vietnamica 145 volume 36, number 2, 2011, pp. 145–167 some properties of orlicz-lorentz spaces ha huy bang, nguyen van hoang